Transcript Slide 1

Chapter 5 Probability in our Daily Lives

Section 5.1: How can Probability Quantify Randomness?

Learning Objectives

1.

2.

3.

4.

5.

6.

Random Phenomena Law of Large Numbers Probability Independent Trials Finding probabilities Types of Probabilities: Relative Frequency and Subjective

Learning Objective 1: Random Phenomena  For random phenomena, the outcome is uncertain  In the short-run, the proportion of times that something happens is highly random  In the long-run, the proportion of times that something happens becomes very predictable Probability quantifies long-run randomness

Learning Objective 2 : Law of Large Numbers  As the number of trials increase, the proportion of occurrences of any given outcome approaches a particular number “in the long run”  For example, as one tosses a die, in the long run 1/6 of the observations will be a 6.

Learning Objective 3: Probability  With random phenomena, the

probability

of a particular outcome is the proportion of times that the outcome would occur in a long run of observations  Example:  When rolling a die, the outcome of “6” has probability = 1/6. In other words, the proportion of times that a 6 would occur in a long run of observations is 1/6.

Learning Objective 4 : Independent Trials  Different trials of a random phenomenon are

independent

if the outcome of any one trial is not affected by the outcome of any other trial.

 Example:  If you have 20 flips of a coin in a row that are “heads”, you are not “due” a “tail” - the probability of a tail on your next flip is still 1/2. The trial of flipping a coin is independent of previous flips.

Learning Objective 5: How can we find Probabilities?

 Calculate theoretical probabilities based on assumptions about the random phenomena. For example, it is often reasonable to assume that outcomes are equally likely such as when flipping a coin, or a rolling a die.

 Observe many trials of the random phenomenon and use the sample proportion of the number of times the outcome occurs as its probability. This is merely an estimate of the actual probability.

Learning Objective 6 : Types of Probability: Relative Frequency vs. Subjective  The

relative frequency definition of probability

is the long run proportion of times that the outcome occurs in a very large number of trials - not always helpful/possible.  When a long run of trials is not feasible, you must rely on subjective information. In this case, the

subjective definition of the probability

of an outcome is your degree of belief that the outcome will occur based on the information available.

 Bayesian statistics is a branch of statistics that uses subjective probability as its foundation

Chapter 5: Probability in Our Daily Lives

Section 5.2: How Can We Find Probabilities?

Learning Objectives

1.

2.

3.

4.

5.

Sample Space Event Probabilities for a sample space Probability of an event Basic rules for finding probabilities about a pair of events

Learning Objectives

6.

7.

Probability of the union of two events Probability of the intersection of two events

Learning Objective 1 : Sample Space  For a random phenomenon, the sample space is the set of all possible outcomes.

Learning Objective 2 : Event  An event is a subset of the sample space  An event corresponds to a particular outcome or a group of possible outcomes.

 For example;  Event A = student answers all 3 questions correctly = (CCC)  Event B = student passes (at least 2 correct) = (CCI, CIC, ICC, CCC)

Learning Objective 3 : Probabilities for a sample space Each outcome in a sample space has a probability  The probability of each individual outcome is between 0 and 1  The total of all the individual probabilities equals 1.

Learning Objective 4: Probability of an Event  The probability of an event A, denoted by P(A), is obtained by adding the probabilities of the individual outcomes in the event.

 When all the possible outcomes are equally likely:

P

(

A

)  number of outcomes in event A number of outcomes in the sample space

Learning Objective 4: Example: What are the Chances of being Audited?

 What is the sample space for selecting a taxpayer?

{(under $25,000, Yes), (under $25,000, No), ($25,000 $49,000, Yes) …}

Learning Objective 4: Example: What are the Chances of being Audited?

For a randomly selected taxpayer in 2002,  What is the probability of an audit?

 310/80200=0.004

 What is the probability of an income of $100,000 or more?

 10700/80200=0.133

 What income level has the greatest probability of being audited?

 $100,000 or more = 80/10700= 0.007

Learning Objective 5: Basic rules for finding probabilities about a pair of events  Some events are expressed as the outcomes that  Are not in some other event (complement of the event)  Are in one event and in another event (intersection of two events)  Are in one event or in another event (union of two events)

Learning Objective 5 : Complement of an event  The complement of an event A consists of all outcomes in the sample space that are not in A.  The probabilities of A and of A c add to 1  P(A c ) = 1 – P(A)

Learning Objective 5: Disjoint events  Two events, A and B, are disjoint if they do not have any common outcomes

Learning Objective 5: Intersection of two events  The intersection of A and B consists of outcomes that are in both A and B

Learning Objective 5: Union of two events  The union of A and B consists of outcomes that are in A or B or in both A and B.

Learning Objective 6: Probability of the Union of Two Events

Addition Rule: For the union of two events, P(A or B) = P(A) + P(B) – P(A and B) If the events are disjoint, P(A and B) = 0, so P(A or B) = P(A) + P(B)

Learning Objective 6 : Example  80.2 million tax payers (80,200 thousand)  Event A = being audited  Event B = income greater than $100,000  P(A and B) = 80/80200=.001

Learning Objective 7: Probability of the Intersection of Two Events Multiplication Rule: For the intersection of two independent events, A and B, P(A and B) = P(A) x P(B)

Learning Objective 7: Example  What is the probability of getting 3 questions correct by guessing?  Probability of guessing correctly is .2

What is the probability that a student answers at least 2 questions correctly?

P(CCC) + P(CCI) + P(CIC) + P(ICC) = 0.008 + 3(0.032) = 0.104

Learning Objective 7 : Assuming independence  Don’t assume that events are independent unless you have given this assumption careful thought and it seems plausible.

Learning Objective 7: Events Often Are Not Independent  Example: A Pop Quiz with 2 Multiple Choice Questions  Data giving the proportions for the actual responses of students in a class Outcome: II IC CI CC Probability: 0.26 0.11 0.05 0.58

Learning Objective 7: Events Often Are Not Independent     Define the events A and B as follows:  A: {first question is answered correctly}  B: {second question is answered correctly} P(A) = P{(CI), (CC)} = 0.05 + 0.58 = 0.63

P(B) = P{(IC), (CC)} = 0.11 + 0.58 = 0.69

P(A and B) = P{(CC)} = 0.58

 If A and B were independent, P(A and B) = P(A) x P(B) = 0.63 x 0.69 = 0.43

Thus, in this case, A and B are not independent!

Chapter 5 Probability in Our Daily Lives

Section 5.3: Conditional Probability: What’s the Probability of A, Given B?

Learning Objectives

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2.

3.

Conditional probability Multiplication rule for finding P(A and B) Independent events defined using conditional probability

Learning Objective 1: Conditional Probability  For events A and B, the conditional probability of event A, given that event B has occurred, is: P(

A

|

B

) 

P

(

A and B

)

P

(

B

)  P(A|B) is read as “the probability of event A, given event B.” The vertical slash represents the word “given”. Of the times that B occurs, P(A|B) is the proportion of times that A also occurs

Learning Objective 1 : Conditional Probability

Learning Objective 1 : Example 1

Learning Objective 1 : Example 1

Learning Objective 1 : Example 1  What was the probability of being audited, given that the income was ≥ $100,000?

 Event A: Taxpayer is audited  Event B: Taxpayer’s income ≥ $100,000 P(A | B)  P(A and B) P(B)  0 .

0010 0 .

1334  0 .

007

Learning Objective 1 : Example 1  What is the probability of being audited given that the income level is < $25,000   Let A =Event being Audited Let B = Income < $25,000 P(

A

|

B

) 

P

(

A

   P(B)=.1758

.0011/.1758=.0063

and P

(

B

) P(A and B) = .0011

B

)

Learning Objective 1: Example 2 

A study of 5282 women aged 35 or over analyzed the Triple Blood Test to test its accuracy

Learning Objective 1: Example 2 

A positive test result states that the condition is present

A negative test result states that the condition is not present

False Positive : Test states the condition is present, but it is actually absent

False Negative : Test states the condition is absent, but it is actually present

Learning Objective 1: Example 2 

Assuming the sample is representative of the population, find the estimated probability of a positive test for a randomly chosen pregnant woman 35 years or older

P(POS) = 1355/5282 = 0.257

Learning Objective 1: Example 2 

Given that the diagnostic test result is positive, find the estimated probability that Down syndrome truly is present

P(D | POS)  P(D and POS) P(POS)  48 / 5282 1355 / 5282  0 .

009 0 .

257  0 .

035 

Summary: Of the women who tested positive, fewer than 4% actually had fetuses with Down syndrome

Learning Objective 2 : Multiplication Rule for Finding P(A and B) 

For events A and B, the probability that A and B both occur equals:

P(A and B) = P(A|B) x P(B) also

P(A and B) = P(B|A) x P(A)

Learning Objective 2: Example 

Roger Federer – 2006 men’s champion in the Wimbledon tennis tournament

He made 56% of his first serves

He faulted on the first serve 44% of the time

Given that he made a fault with his first serve, he made a fault on his second serve only 2% of the time

Learning Objective 2: Example 

Assuming these are typical of his serving performance, when he serves, what is the probability that he makes a double fault?

P(F1) = 0.44

P(F2|F1) = 0.02

P(F1 and F2) = P(F2|F1) x P(F1) = 0.02 x 0.44 = 0.009

Learning Objective 3 : Independent Events Defined Using Conditional Probabilities 

Two events A and B are independent if the probability that one occurs is not affected by whether or not the other event occurs

Events A and B are independent if: P(A|B) = P(A), or equivalently, P(B|A) = P(B)

If events A and B are independent, P(A and B) = P(A) x P(B)

Learning Objective 3 : Checking for Independence 

To determine whether events A and B are independent:

Is P(A|B) = P(A)?

Is P(B|A) = P(B)?

Is P(A and B) = P(A) x P(B)?

If any of these is true, the others are also true and the events A and B are independent

Learning Objective 3: Example 

The diagnostic blood test for Down syndrome: POS = positive result NEG = negative result D = Down Syndrome D C = Unaffected Status POS NEG D D c 0.009

0.247

0.001

0.742

Total 0.257

0.743

Total 0.010

0.990

1.000

Learning Objective 3: Example: Checking for Independent 

Are the events POS and D independent or dependent? Is P(POS|D) = P(POS)?

P(POS|D) =P(POS and D)/P(D) = 0.009/0.010 = 0.90

P(POS) = 0.257

The events POS and D are dependent

Chapter 5: Probability in Our Daily Lives

Section 5.4: Applying the Probability Rules

Learning Objectives

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2.

3.

4.

Is a “Coincidence” Truly an Unusual Event?

Probability Model Probabilities and Diagnostic Testing Simulation

Learning Objective 1 : Is a “Coincidence” Truly an Unusual Event?

The law of very large numbers states that if something has a very large number of opportunities to happen, occasionally it will happen, even if it seems highly unusual

Learning Objective 1: Example: Is a Matching Birthday Surprising?

What is the probability that at least two students in a group of 25 students have the same birthday?

Learning Objective 1: Example: Is a Matching Birthday Surprising?

P(at least one match) = 1 – P(no matches)

Learning Objective 1: Example: Is a Matching Birthday Surprising?

P(no matches) = P(students 1 and 2 and 3 …and 25 have different birthdays)

Learning Objective 1: Example: Is a Matching Birthday Surprising?

P(no matches) = (365/365) x (364/365) x (363/365) x … x (341/365)

P(no matches) = 0.43

Learning Objective 1: Example: Is a Matching Birthday Surprising?

P(at least one match) = 1 – P(no matches) = 1 – 0.43 = 0.57

Not so surprising when you consider that there are 300 pairs of students who can share the same birthday!

Learning Objective 2: Probability Model

 We’ve dealt with finding probabilities in many idealized situations  In practice, it’s difficult to tell when outcomes are equally likely or events are independent  In most cases, we must specify a

probability model

that approximates reality

Learning Objective 2: Probability Model

 A probability model specifies the possible outcomes for a sample space and provides assumptions on which the probability calculations for events composed of these outcomes are based  Probability models merely approximate reality

Learning Objective 2: Example: Probability Model

 Out of the first 113 space shuttle missions there were two failures  What is the probability of at least one failure in a total of 100 missions?

 P(at least 1 failure)=1-P(0 failures) =1 P(S1 and S2 and S3 … and S100) =1 P(S1)xP(S2)x…xP(S100) =1-[P(S)] 100 =1-[0.971] 100 =0.947

Learning Objective 2: Example: Probability Model

 This answer relies on the assumptions of  Same success probability on each flight  Independence These assumptions are suspect since other variables (temperature at launch, crew experience, age of craft, etc.) could affect the probability

Learning Objective 3: Probabilities and Diagnostic Testing

Sensitivity

= P(POS|S) 

Specificity

= P(NEG|S C )

Learning Objective 3: Example: Probabilities and Diagnostic Testing Random Drug Testing of Air Traffic Controllers  Sensitivity of test = 0.96

 Specificity of test = 0.93

 Probability of drug use at a given time ≈ 0.007 (prevalence of the drug)

Learning Objective 3: Example: Probabilities and Diagnostic Testing What is the probability of a positive test result?

P(POS)=P(S and POS)+P(S C and POS)  P(S and POS)=P(S)P(POS|S) = 0.007x0.96=0.0067

 P(S C and POS)=P(S C )P(POS|S C ) = 0.993x0.07=0.0695

 P(POS)=.0067+.0695=0.0762

Even though the prevalence is < 1%, there is an almost 8% chance of the test suggesting drug use!

Learning Objective 4: Simulation Some probabilities are very difficult to find with ordinary reasoning. In such cases, we can approximate an answer by

simulation

.

Learning Objective 4: Simulation

Carrying out a Simulation:  Identify the random phenomenon to be simulated  Describe how to simulate observations  Carry out the simulation many times (at least 1000 times)  Summarize results and state the conclusion